leetcode Construct Binary Tree from Preorder and Inorder Traversal
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Construct Binary Tree from Preorder and Inorder Traversal
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
http://blog.csdn.net/kenden23/article/details/15500733
这次是使用preorder数组定位跟节点,利用inorder数组分左右子树。
关键点:
1 定位每层的根节点
2 计算好offset
//2014-2-16 updateTreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {return build(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);}TreeNode *build(vector<int> &preorder, int pre1, int pre2,vector<int> &inorder, int in1, int in2){if (pre1 > pre2) return nullptr;TreeNode *root = new TreeNode(preorder[pre1]);int offset = 0;for ( ; inorder[in1+offset] != preorder[pre1]; offset++);root->left = build(preorder, pre1+1, pre1+offset, inorder, in1, in1+offset-1);root->right = build(preorder, pre1+offset+1, pre2, inorder, in1+offset+1, in2);return root;}
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